All the lepton masses from G, pi, e

[CarlB]

After the pion, the next most basic meson is the rho. There are 5 rho resonances, the rho(770), rho(1450), rho(1700), rho(1900), and rho(2150). The last two are "omitted from the summary section" by the Particle Data Group. The lower three can be put into Koide form as follows:

\lambda_{\rho n} = 25.0544\sqrt{\textrm{MeV}}(10/7 -1/3 \cos(2/9 + 2n\pi/3)\;)

This is fairly similar to the formula for the pion:
\lambda_{\pi n} = 25.0544\sqrt{\textrm{MeV}}(6/5 -3/4 \cos(2/9 + 2n\pi/3)\;)

And of course the Koide formula uses the same angle 2/9.

I don't feel very comfortable about the rational values of the v = 10/7, 6/5, and s= 1/3 and 3/4 numbers. I feel better about the fact that v+s/2 tends to be constant when you are looking at two different resonance series. That is, 10/7 + (1/3)/2 is about equal to 6/5 + (3/4)/2 to 1%. There's a good example of this 2 to 1 ratio in some of the longer resonance series where you have more terms in the series.

The reason I like this 2 to 1 relationship between v and s is because in the derivation of the previous post, it comes up with v + 2 s \cos(\delta + 2n\pi/3), so changing v and s by 2 to 1 means that the vector length is split between the constant part (the valence part) "v", and the variable or sea part "s".

I'm still thinking on Hans' equation. The kind of thing I'm looking at is geometric as in:
http://www.sparknotes.com/math/geometry2/theorems/section5.rhtml

 

[Kea]

After the pion, the next most basic meson is the rho...

Carl, of course it is nice to have a derivation of so many particle masses, but I am guessing that people will still complain that v and s constitute 2 parameters for 2 outcomes in each case (forgetting for the moment the nice 'coincidence' of choice of scale etc), so could you please elaborate a little on the field idea, and the choice of these variables.

 

[CarlB]

Well Kea, there are three masses so three degrees of freedom. The "v" and "s" only provide two degrees of freedom. The coincidence is in the 2/9. I'm working on a nice graphic that will show how the coincidence works, for inclusion in the paper, which I guess I ought to give the first cut on:
http://www.brannenworks.com/qbs.pdf

The idea is to plot the angle delta as a function of the power one chooses to take. That is to explore mass formulas that look like:

m^{k} = v + s\cos(\delta + 2n\pi/3)

where the three parameters are v, s, and delta, and to plot delta as a function of k. I believe that a nice coincidence will show up when k=1/2, that is the curves for the various mesons the approrpriate baryons, and the leptons will all converge to 2/9 at that value of k.

Carl

 

[arivero]

I think that one must consider some logical grouping for the mesons. For instance, when we recorded the sqrt(m1) sqrt(m2-m3) permutations, we related them to the level spliting of the flavour octet.

 

[CarlB]

Here's a mass formula for the pions that I hope the reader will find amusing. Begin with the mass formula for the charged leptons (since I want to use "n" for radial excitations of mesons, I will use "g" for the quantum number that gives the generation):

\sqrt{m_{e g}}/25.0544 = \sqrt{0.5} + \cos(2/9 + 2g\pi/3)

where 25.0544 \sqrt{\textrm{eV}} is a mass scaling constant that makes the leptons nice and we will use for the mesons as well.

The basic idea is to think of the mesons as having radial excitations "n" with quantum numbers like the spherical harmonics of the hydrogen atom, but also having color excitations following Koide's formula. This means that we get the hydrogen wave functions, but tripled.

Guess that the three lowest mass pions = (\pi, \pi(1300), \pi(1800)) with masses (139.57, 1300, 1812) are the n=1, l=0 states with three Koide generations distinguishing them. Let's use "g" for the generation number. Then the formula for these three pion masses turns out to be:

\sqrt{m_{\pi 10g}}/25.0544 = 1.196797 -0.743543\cos(2/9 + 2g\pi/3).

Make the same guess about the three lowest mass J=1 pions = (\pi_1(1400), \pi_1(1600), \pi_1(2015)). Their masses are (1376, 1653, 2013). Since J=1, these have n=2, l=1. Note the last of these, along with the last two in the next group of three, is hard to find in the PDG. It's listed on the page titled "further states":
http://pdg.lbl.gov/2007/listings/m300.pdf
The Koide formula for these j=1 pion masses is:

\sqrt{m_{\pi 21g}}/25.0544 = 1.6313 + 0.1792986\cos(2/9 + 2g\pi/3 + \pi/12).

The extra angle pi/12 means that these have a formula like the neutrinos.

And the three lowest J=2 pions are the (\pi_2(1670), \pi_2(1880), \pi_2(2005)) with masses (1672.4, 1880, 1990). These presumably have n=3, l=2. They also have a Koide formula:

\sqrt{m_{\pi 32g}}/25.0544 = 1.71477 + 0.0864566\cos(2/9 + 2g\pi/3 + 2\pi/12).

The above 3 equations give 9 masses and they work pretty well.

For the hydrogen excitations, the total energy (or mass) of the hydrogen atom is approximately:
m_{H n} = 10^9 - 13.6 /n^2 eV.
The second term, the binding energy, is very small compared to the total energy of the hydrogen atom (about 1GeV) which is mostly due to the rest mass of the proton. Consequently, if I rewrite this as a formula for the square root of the mass of the hydrogen atom, the square root energy levels will still follow a 1/n^2 law:
\sqrt{m_{H n}} = 10^{9/2} - (10^{-9/2}\times 13.6/2) /n^2 \sqrt{eV}.

So based on the energies of the hydrogen atom, one might look for 1/n^2 dependency in the three Koide mass formulas given above, in order to unify the three equations. The three "v" terms need to be: (1.19680, 1.6313, 1.71477). These have n=(1,2,3). They can be fairly well approximated with the formula v_n = 16/9-\sqrt{1/3}/n^2. The three "s" terms need to be (-0.743543, +0.1792986, +0.0864566 ). The sign change can be accounted for by making all three signs negative, but taking an extra phase of pi for the n=1 and n=2 case. This can be accomplished by adding a phase of nj\pi/2=n(n-1)\pi/2 to the phase angle. Then the numbers are fairly closely approximated by s_n = -0.75/n^2. The resulting formula for nine pion masses is fairly compact:

\sqrt{m_{\pi nlg}}/25.0544 = 16/9 - \sqrt{3}/n^2 -(3/4)\cos(2/9 + 2n\pi/3 + jn\pi/2)/n^2

The above simplifications give an approximations of the v's as follows:
1.19680 -> 1.2004275
1.6313 -> 1.63344
1.71477 -> 1.71362
And the s's are approximated by:
-0.743543 -> -3/4
+0.1792986 -> 0.1875,
+0.0864566 -> 0.08333
which is fairly close.

The nine computed masses are as follows:
137.99 1271.2 1834.1
1365.5 1659.6 2032.4
1676.1 1883.0 1977.3

The nine actual masses are:
136.5(2.5) 1300(100) 1812(14)
1376(17) 1652(21) 2014(?)
1672.4(3.2) 1876-2003(?) 1974-2005(?)

Note: The error in the pi mass is due to the mass difference between the pi+ and pi0. And the masses labeled with (?) do not have mass spreads in the PDG. If more than one measurement is given, the range of measurements are given.

 

[CarlB]

Ooops. Correction on that mass formula. Managed to make three errors in it. (And what happened to the edit function within 24 hours?)

\sqrt{m_{\pi nlg}}/25.0544 = 16/9 - \sqrt{3}/n^2 -(3/4)\cos(2/9 + 2g\pi/3 + l\pi/12 + ln\pi/2)/n^2

In the above, for the pion states it applies to, l is assumed to be equal to n-1. The full set of quantum numbers is nlmg rather than nlg, but there is no energy dependency on m.

And I guess I should note that 16/9 = 2 - 2/9. So the valence part of the formula can be interpreted as the contribution from the seas of the two quarks (whose valence parts have canceled because one is a quark and the other is an anti-quark), and the 2/9 may be related to the 2/9 in the cosine.

I've got a lot more of these. One of the cool ones to analyze is the J/Psi. There are 6 states with masses fairly close together. Is there a way of dividing them up into two groups of three, where each group of three fits the above kind of formula? Let's see if I can get a solution edited into this post before it times me out.

 

[CarlB]

Remember Foot's gemoetric interpretation of Koide's formula in terms of the vector (1,1,1)? It turns out that the same may be said of the tribimaximal mixing matrix. This should be of interest to those interested in Koide's formula:
http://carlbrannen.wordpress.com/2008/06/29/the-mns-matrix-as-magic-square/

Maybe we will finally get a complete Koide type formulation for the elementary particles as a similar property applies to the CKM matrix as well. (That is, Kea is says that the CKM matrix can be written as a sum of a 1-circulant and 2-circulant matrix, and this is related to the Foot geometry, as shown above.)

Also, I should mention that I've got good results applying the Koide formula to the b-bbar and c-cbar mesons. It could be an interesting summer.

 

[arivero]

2008 review

CITATION: C. Amsler et al. (Particle Data Group), Phys. Lett. B667, 1 (2008) has some changes in W mass and width, as well as others.


The strange news: [/B]

with the move of W mass from 80.425 down to 80.398 ± 0.025 GeV the fit of me/mW to the muon anomalous difference, in post #41, should be worse. But the pdg now lists .62686, an increased difference. So both moves sort of cancel.

the positive review:

Alpha formula, in post #4, survives. But term 4 should be included in next revisions!
after term 3: 0.0072973525686533
f.s. constant 0.0072973525680.(+/-240) old
f.s. constant 0.0072973525692 .(+/-27) http://arxiv.org/abs/0712.2607

post #28, Pythagorean triples, would go better or similar, with actual tau mass.

#44, Weinberg angle, enters the 1-sigma! In 2004 it was to within 0.063% or sigma 1.2). Now with the 2008 pdg, it predicts W mass within 0.029% and sigma 0.94

The prediction is more spectacular if you see post #66, column "calculated in MeV"

 

[arivero]

Some works are borderline between standard "texture" research and our thread.
In http://arxiv.org/pdf/hep-ph/9703217v1, they suggest
<br /> \sqrt {\m_t\over\m_c} = {m_\tau\over m_\mu}<br /> [\tex]<br /> In <a href="http://arxiv.org/pdf/hep-ph/0106286v2" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://arxiv.org/pdf/hep-ph/0106286v2</a>, a lot of such sqrt textures are invoked as popularly known.<br /> In http://ccdb4fs.kek.jp/cgi-bin/img/allpdf?198812215 and http://dx.doi.org/10.1016/0370-2693(87)91621-2 Christof Wetterich leads other attempts.

 

[arivero]

Alpha formula, in post #4, survives. But term 4 should be included in next revisions!
after term 3: 0.0072973525686533
f.s. constant 0.0072973525680.(+/-240) old
f.s. constant 0.0072973525692 .(+/-27) http://arxiv.org/abs/0712.2607

Correction, the full calculation is in http://www.physics-quest.org/, so we can compare
0.00729735256865385342269 theoretical
0.0072973525692 .(+/-27) measured http://arxiv.org/abs/0712.2607
quotient:

1.000000000075 \pm 0.000000000369

 

[Hans de Vries]

Correction, the full calculation is in http://www.physics-quest.org/, so we can compare
0.00729735256865385342269 theoretical
0.0072973525692 .(+/-27) measured http://arxiv.org/abs/0712.2607
quotient:

1.000000000075 \pm 0.000000000369

Wow!

That's a nice surprice! New measurements plus a revised value of the eighth-order
QED contribution to the anomalous magnetic moment of the electron from Kinogarbagea
brings our series in line with experiment to a fraction of the (improved) error.

\alpha^{-1}~=~~ 137.035 999 710 (96) Old Kinogarbagea/Gabrielse (July 2006)
\alpha^{-1}~=~~ 137.035 999 084 (51) New Kinogarbagea/Gabrielse (Feb 2008)
\alpha^{-1}~=~~ 137.035 999 095 829 Theoretical value from alpha series.

New measurement from Gabrielse's group: http://arxiv.org/abs/arXiv:0801.1134v1
Revision of the eight-order QED term: http://arxiv.org/abs/0712.2607
The alpha series: http://physics-quest.org/fine_structure_constant.pdf

This happened a while ago already. somehow we missed it. Thank you for posting!



Regards, Hans.

 

[arivero]

To complete the record:

\alpha^{-1}~=~~ 137.035 999 108 (450) Original conject. from Codata 2004.
\alpha^{-1}~=~~ 137.035 999 710 (96) Old Kinogarbagea/Gabrielse (July 2006)
\alpha^{-1}~=~~ 137.035 999 084 (51) New Kinogarbagea/Gabrielse (Feb 2008)

\alpha^{-1}~=~~ 137.035 999 095 829 Theoretical value from alpha series.

Wow!
This happened a while ago already. somehow we missed it. Thank you for posting!

I somehow fused it with the already reported, somewhere in the middle of the thread, 2006 update.

As for the calculation, the first order formula
<br /> \alpha^{-1/2}+ \alpha^{1/2}=e^{\pi^2 \over 4}<br />

is crying a word: duality. Problems are:
1) It is not so clear how the succesive corrections are applied.
2) the formula for dyon energy (particle with electric+monopole charge) is
<br /> M^2 \approx e^2 + e^{-2} = \alpha^{1}+ \alpha^{-1}<br />
so this formula is a kind of rare square root of the usual duality formula.

3)Still we haven't got a clue for the precise choosing of e(pi^2/4) except that it works.

 

[Hans de Vries]

To complete the record:

\alpha^{-1}~=~~ 137.035 999 110 (450) Original conject. from Codata 2004.
\alpha^{-1}~=~~ 137.035 999 710 (96) Old Kinogarbagea/Gabrielse (July 2006)
\alpha^{-1}~=~~ 137.035 999 084 (51) New Kinogarbagea/Gabrielse (Feb 2008)

\alpha^{-1}~=~~ 137.035 999 095 829 Theoretical value from alpha series.

Both the direct measurement from the Quantum Hall effect (CODATA 2004) and
the indirect one from the Harvard g/2 measurements are now in agreement.

As for the calculation, the first order formula
<br /> \alpha^{-1/2}+ \alpha^{1/2}=e^{\pi^2 \over 4}<br />

is crying a word: duality. Problems are:
1) It is not so clear how the succesive corrections are applied.
2) the formula for dyon energy (particle with electric+monopole charge) is
<br /> M^2 \approx e^2 + e^{-2} = \alpha^{1}+ \alpha^{-1}<br />
so this formula is a kind of rare square root of the usual duality formula.

3)Still we haven't got a clue for the precise choosing of e(pi^2/4) except that it works.

To explain it is the hard thing... Do you have a link for the this dyon
mass formula?Regards, Hans

 

[arivero]

To explain it is the hard thing... Do you have a link for the this dyon
mass formula?

In fact, the point of not having recognised it immediately is proof of the amateurish character of the internet forums

Remember the lore: if a theory goes with a coupling constant a, the dual theory goes usually with a coupling constant 1/a.

In the case of T-duality the coupling have dimensions. Or said otherwise, Tduality is possible because the theory has a dimensional scale. But for ElectroMagnetic duality, the coupling is adimensional. The (square of the?) magnetic energy of a monopole goes as hbar/alpha.

Actually, a lot of the initial formulae you suggested in 2004 were of the form x+1/x for some constant x; it is hard of believe that you were not inspired, at least inconsciently, by the usual popular remarks on duality. Again, I failed to notice it, so perhaps both did.

A dyon is a particle having both elemental and monopole charges. So its mass square has two contributions.

I saw the formula for the mass in a remark about http://www.slac.stanford.edu/spires/find/hep/www?irn=251658"

The dyon mass formula appears in the last page of Montonen-Olive preprint, and they quote towards to older articles.

To put some stuff in handwritten from Witten itself, see slide 25 of http://math.berkeley.edu/index.php?...tManager_op=downloadFile&JAS_Document_id=2101 in http://math.berkeley.edu/index.php?...ntManager_op=viewDocument&JAS_Document_id=116
and then slide 28

 

[arivero]

Perhaps I have cited very heavy artillery. I apologize; I have never put a lot of interest on these topological objects; it is a long history.

So another source is Zee's book on QFT, and it redirects to http://www.maths.ed.ac.uk/~jmf/Teaching/EDC.html" , from Figueroa-o'Farrill.

In section 1.3.1 we are told that the mass of a monopole always meets the Bogomol’nyi bound. In section 1.3.2 the bound is "saturated" in the Prasad–Sommerfield limit. So we have the initials BPS. The saturated bound is the formula for the mass of a monopole with both electric and magnetic charge I told above, depending unfortunately of \alpha^1 instead of \sqrt{\alpha} \; (=e).
Alert: Section 1.4.2 explains another earlier calculation of Witten, and it builds the "Noether" charge of a dyon as funtion of charges q g plus a theta-vacuum, in the shape
<br /> N={q \over e} + (e g) {\theta \over 8 \pi^2}<br />
and then argues that N must be an integer. But note that q and g are not fully adimensional; they are defined as, say q=(n e) and (g=4 pi m/e), so the factors of e there in the formula are intended only to cancel the internal ones. The argument in 1.4.2 amounts to prove that when m=1, n is also quantized.

 

[arivero]

Could it be better to write the first order approximation squared, ie

<br /> e^{\pi^2/2} -2 \approx \alpha + \alpha^{-1}<br />

In such case, how could the corrective terms be applied?

 

[CarlB]

Or is it useful to begin with a first order equation:
\sqrt{\alpha} = i^{\ln(i)} = i^{i\pi/2}
= (i^i)^{\pi/2} = (e^{-\pi/2})^{\pi/2} = e^{-\pi^2/4}

Or did I make an algebra mistake here?

 

[Kea]

Assuming that the first order relation is indicative of duality, this would be Langlands, or electric magnetic duality, most simply understood via relations coming from the modular group, or its braid group cover. This suggests interpreting the Gaussian term as a trace factor (usually called d in knotty algebra maths), or a normalisation arising from geometry of non integral dimension, not unlike Deligne's Gaussian for the discrete Fourier transform, except that the factor of \pi presumably results from an infinite sequence of terms.

 

[arivero]

Or is it useful to begin with a first order equation:
\sqrt{\alpha} = i^{\ln(i)} = i^{i\pi/2}
= (i^i)^{\pi/2} = (e^{-\pi/2})^{\pi/2} = e^{-\pi^2/4}

Or did I make an algebra mistake here?

No, I think it is fine. Of course i=e^{i\pi /2} is a nice trick to get this power of pi. Actually it shows that the seed chosen by Hans is not a random number.

Most important perhaps, you can -as Hans hinted in 2004, calling this a "gaussian"- also write in a formal way

<br /> {i \over \sqrt \pi} \int_{-\infty}^\infty e^{ (x-0) (x\pm \pi)} dx = e^{- {\pi^2 / 4}}<br /> or if you prefer <br /> {1 \over \sqrt \pi} \int_{-\infty}^\infty e^{ - (x-0) (x\pm i \pi)} dx <br />

it is because of this second detail that I have wasted, poorly, the weekend looking into soliton, instanton, dyons etc...ons. Not only the x-(1/x) part of the equality is typical of this kind of theories; also the integral of gaussians is the part to get exact values of energy, barrier penetration, mass etc. So the "1-st order" (Hans call the one you wrote "0-th order") formula has common points with non-perturbative QM/QFT/Qthing in both sides of the expression, which is intriguing.

 

[Count Iblis]

You only got ten digits, that's not enough to guess the correct formula.

 

[Hans de Vries]

I'm sure the magnetic moment plays a major role, which reminds me of another electro-
magnetic duality: There are types of two magnetic dipoles with indistinguishable fields,
the vector dipole and the axial dipole. The first is a combination of two opposite magnetic
monopoles and the second is a point like circular electric current.

So, associating the field with a continuous charge/spin distribution, which of the two
types is involved in case of the magnetic moment? Well, the net force in a magnetic
field is quite different:

Force on a Vector dipole in a B field:

<br /> \begin{array}{l l |l l l| l | l | l}<br /> &amp; &amp; \partial_x \textsf{B}_x &amp; \partial_y \textsf{B}_x &amp; \partial_z \textsf{B}_x &amp; &amp; \mu_x &amp; \\<br /> \vec{F}_{magn} &amp; =\ \ &amp; \partial_x \textsf{B}_y &amp; \partial_y \textsf{B}_y &amp; \partial_z \textsf{B}_y &amp; \cdot &amp; \mu_y &amp; \quad =\ \partial_j B_i\ \mu_i \\<br /> &amp; &amp; \partial_x \textsf{B}_z &amp; \partial_y \textsf{B}_z &amp; \partial_z \textsf{B}_z &amp; &amp; \mu_z<br /> \end{array}<br />

Force on an Axial dipole in a B field:

<br /> \begin{array}{l l |l l l| l | l | l}<br /> &amp; &amp; \partial_x \textsf{B}_x &amp; \partial_x \textsf{B}_y &amp; \partial_x <br /> \textsf{B}_z &amp; &amp; \mu_x &amp; \\<br /> \vec{F}_{magn} &amp; =\ \ &amp; \partial_y \textsf{B}_x &amp; \partial_y \textsf{B}_y &amp; \partial_y \textsf{B}_z &amp; \cdot &amp; \mu_y &amp; \quad =\ \partial_i B_j\ \mu_i \\<br /> &amp; &amp; \partial_z \textsf{B}_x &amp; \partial_z \textsf{B}_y &amp; \partial_z \textsf{B}_z &amp; &amp; \mu_z<br /> <br /> \end{array}<br />

Note the exchange of the indices. The second case is what you would expect:

\vec{F} ~~=~~\mbox{grad}(\vec{B}\cdot\vec{\mu})


So, it should be easy to distinguish between the two cases you would say,
and this rules out the pair of magnetic monopoles. There is an interesting twist
however: The net force between either two vector dipoles or two axial dipoles
turns out to be same:

<br /> \vec{F}\ =\ \frac{3\mu_o\mu_e^2}{4\pi r^4}\left[ \left\{\left(<br /> \hat{\mu}_1 \cdot \hat{\mu}_2 \right) - 5\left( \hat{r} \cdot<br /> \hat{\mu}_1 \right)\left( \hat{r} \cdot \hat{\mu}_2 \right)\right\}<br /> \mbox{\Large $\hat{r}$} + \left( \hat{r} \cdot \hat{\mu}_2<br /> \right)\mbox{\large $\hat{\mu}_1$} + \left( \hat{r} \cdot<br /> \hat{\mu}_1 \right)\mbox{\large $\hat{\mu}_2$}\ \right]<br />

At least in the case that both are at rest, which reminds me that I should find some
time to work out the general case too really proof the exclusion of the first case.


Regards, Hans

 

[CarlB]

Okay, if the algebra works out, (it is usual for me to get 2s and factors of pi wrong the first time), then I should give credit where credit is due. The formula for i^i was discussed at Built On Facts:
http://scienceblogs.com/builtonfacts/2008/08/sunday_function_4.php

So what are the attributes of the complex mapping
f(u) = u^{\ln(u)}
and why would you be interested in this function at the point i?

What are the poles of ln(u)?

I bet Kea can tie this into the Riemann zeta function.

 

[arivero]

Just to fix the idea, the point about the exp is that it is gaussian but a peculiar one. It is a wavepacket with the momenta distribution centered in pi instead of the origin. To be precise, the normalized solution of schroedinger equation, at t=0, for a gaussian packet of mean wavenumber k_0 and distribution width a=1.
<br /> \Psi(x)= ({2\over \pi})^{1/4} e^{i k_0 x} e^{-x^2}<br />
so that at least formally
<br /> \int_{-\infty}^{\infty} \Psi(x) dx = ({2 \pi})^{1/4} e^{-k_0^2\over 4} <br />

And NOW I see that I do not understand why I am trying to fit the minus sign; Hans "first order" formula uses i^(-ln i). CarlB reversed it only to use the 0th-order.

So it is the reverse situation: we actually have a wavepacket of imaginary average wavenumber k_0=i \pi. I'd say that this kind of beast fits with the definition of instanton.

 

[arivero]

An extra rewrite: \sqrt \alpha= e /\sqrt {2 \pi} so

e + {2 \pi \over e} = \sqrt{2 \pi} \rm e^{\pi^2/4}

(typography problem: should we use \rm for the exponential or for the electron charge?)

While it seems lot uglier, it agrees with the conventions for Dirac monopoles, whose charge g is such that
e g = 2 \pi n
for n any integer.

 

[arivero]

So, associating the field with a continuous charge/spin distribution, which of the two
types is involved in case of the magnetic moment? Well, the net force in a magnetic
field is quite different:

Witten's http://ccdb4fs.kek.jp/cgi-bin/img_index?197909065
quotes some old (1936-1949) works on "heuristic derivations" of the relationship between electric and magnetic charge when both kinds are present in the same point.

Thoughts of today, if the exponential is actually a gaussian wavepacket:

- there is a imaginary momentum, and this is Euclidan wick rotation.
- such momentum is a multiple of pi, this is periodicity or circular configuration: U(1)??
- self duality is usually the hallmark of a BPS state.

the problem is how has a field theoretical beast, the electromagnetic instanton, descended to the realm of naive quantum mechanics. Of course we "known" the answer: it can only descend for a particular value of alpha. But we do not know what the question is.

 

[CarlB]

the problem is how has a field theoretical beast, the electromagnetic instanton, descended to the realm of naive quantum mechanics. Of course we "known" the answer: it can only descend for a particular value of alpha. But we do not know what the question is.

Hmm, interesting that you would put it that way. All the Koide formulas seem to be reduction of a QFT problem to QM methods.

And the paper I'm working on with the hadron mass formulas amounts to the same thing.

Naive QM wise, we can think of the meson as one quark moving in the field of the other, that is, we go to center of mass coordinates in both space and color. As a first approximation, assume the quark is in a 1S state and we will ignore spatial degrees of freedom. What's left is color degrees of freedom, R, G, and B.

Let H be the Hamiltonian for the meson. Since there's no spatial dependency, H is only a 3x3 matrix, with color indices:
H = \left(\begin{array}{ccc}<br /> H_{RR}&amp;H_{RG}&amp;H_{RB}\\<br /> H_{GR}&amp;H_{GG}&amp;H_{GB}\\<br /> H_{BR}&amp;H_{BG}&amp;H_{BB}\end{array}\right)
By color symmetry,
H_{RR} = H_{GG} = H_{BB} = v,
H_{RG} = H_{GB} = H_{BR} = se^{i\delta},
H_{RB} = H_{GR} = H_{BG} = s&#039;e^{i\delta&#039;}.

To get real eigenvalues H must be Hermitian so v is real, s=s' is real, and \delta&#039; = -\delta. The three eigenvectors are (1,1,1), (1,w,w*), (1,w*,w), where w is the cubed root of unity, and the three eigenvalues are:
s+2v\cos(\delta + 2n\pi/3),
for n = 0,1,2.

The above is almost in the form of Koide's formula. The difference is that Koide's formula is for sqrt(H) instead of H. To get that last step, note that, without color, the non relativistic Hamiltonian is in the form:
H = \nabla^2 + V
where V is a potential, and a slightly more complicated equation for relativistic energy. To get this into clean form, we do the same thing Dirac did to get the gamma matrices, that is, we take the square root. The difference is that in our case, we are taking the square root of a 3x3 color matrix instead of the square root of a d'Alembertian (or whatever the 4-d gradient is called).

I'm working on the statistics of the fits for 20 hadron triplets to this formula and should get something done in the next few days. I've discussed various fits here and elsewhere, but this wil be the first time that everything is brought into one paper, along with the statistical fits. This involves a fair amount of computer programming.

 

[arivero]

Hmm, interesting that you would put it that way. All the Koide formulas seem to be reduction of a QFT problem to QM methods.

And the paper I'm working on with the hadron mass formulas amounts to the same thing.

Yes, and Hans' version of the Weinberg angle was also a QM object. In general, this thread is defying the lore of the origin of constants from HEP (from Planck scale GUT groups). Of course, if we don't go high (on energy, I mean) we do not need to create/annihilate particles, and QM should work.

What worries me today is that the we are not looking in a far dark corner; your angles on neutrinos were a "minor" mainstream topic in recent years. The reinterpretation of Hans alpha as coming from self dual objects is not a minor stream, it is a major fluent of the Amazon river. Damn, it is just a "wrong sign version" of expression 1.2 of hep-th/9407087.

 

[Kea]

What worries me today is that the we are not looking in a far dark corner...

Why is that a concern? Are you worried that the string theorists will swallow this whole? If they had really understood these kind of observations, we would have discovered it by now.

 

[arivero]

Why is that a concern? Are you worried that the string theorists will swallow this whole? If they had really understood these kind of observations, we would have discovered it by now.

More or less, this is the point. It is hard to think that they can overlook a critical value of alpha in a theory whose main paper has almost two thousand citations. They should have discovered it by now. The gate is not very hidden; probably it amounts to consider f(e+g) instead of f(e+ig) as they usually do. And there are some hints that they have tried, notably Poliakov and a lot of stuff on condensates.

 

[arivero]

I'm sure the magnetic moment plays a major role, which reminds me of another electro-
magnetic duality: There are types of two magnetic dipoles with indistinguishable fields,
the vector dipole and the axial dipole. The first is a combination of two opposite magnetic
monopoles and the second is a point like circular electric current.

And I guess you can also compare circular monopole currents against electric charges.

 

[arivero]

Or
<br /> e^{\pi^2/4} - e^{-\pi^2/4} \approx \sqrt{ \alpha + {1 \over \alpha}}<br />
instead of
<br /> e^{\pi^2/4} \approx (\sqrt \alpha + {1 \over \sqrt\alpha})<br />

Actually it is not so good as starting point
11.706956417... / 11.7065492967 (22) = 1.000034777
11.791761389... / 11.7916621597 (22) = 1.000008415
and a sinh is less atractive, to me, than a clean gaussian. And I doubt it could receive the same corrective series than the original guess. It is just that it is closer to popular shapes.

 

[Hans de Vries]

At least the new 2008 value for alpha from Gabrielse/Kinogarbagea begins to
solidify the n=3 term which now gives some convidence in the whole series.

\alpha\ =\ 1/137.035999084 (51)

The alpha "radiative" series:

\mbox{\Huge i}^{~ln\, i }\sum_{n=0}\frac{\mbox{\Huge $\alpha$}^{n-\frac{1}{2}}}{(2\pi)^{b_n}} ~~ = ~~ \mbox{\Huge 1}

Where b_n is the binominal series 0,0,1,3,6,10.. with successive increments
of 0,1,2,3,4...

The result of the series after each extra term is:

n=0: ... 0.992 747 158 626 634
n=1: ... 0.999 991 584 655 288
n=2: ... 0.999 999 998 402 186
n=3: ... 0.999 999 999 957 418
n=4: ... 0.999 999 999 957 464
Regards, Hans

 

[CarlB]

Re: string theory, writing down simple equations like this puts a crimp in the anthropic theories.

 

[jal]

I apologize if I'm am interupting the flow with an irelevant question.
(I can barely understand what you are doing.)
From what you are doing,

arivero,
Just to fix the idea, the point about the exp is that it is gaussian but a peculiar one. It is a wavepacket with the momenta distribution centered in pi instead of the origin. To be precise, the normalized solution of schroedinger equation, at t=0, for a gaussian packet of mean wavenumber k_0 and distribution width a=1.

Can the numbers of wavepacket that are on the shell be calculated?
jal

 

[CarlB]

Getting things to work at low energies instead of the traditional perturbation theory.

I wonder if recent papers by Marco Frasca are related:
http://marcofrasca.wordpress.com/2008/09/03/a-stunning-summer/

 

[arivero]

Re: string theory, writing down simple equations like this puts a crimp in the anthropic theories.

It puts a crimp in GUT almost. The thread is telling that no yukawa couplings are to be predicted from GUT. Moreover, Weinberg angle and fine structure constant comes from Hans, so the "GUT coincidence" can only predict the SU(3) coupling. Furthermore, there is probably a high-low consistence rule: one climbs up from the electroweak scale, get the SU(2) and U(1) couplings to meet, then climbs down the SU(3) coupling until it becomes large and all the QCD nonperturbative stuff (as Marco's) applies, and then surprise, after all this walk we are in a energy scale no far from the original one.

(Of course, it is not fair in a thread on coincidences to discard the One of GUT coupling constants).

Can the numbers of wavepacket that are on the shell be calculated?
jal

Yep, when the packet halfwidth is taken as 1, the wavenumber is k0= i pi, or perhaps i pi/4 depending on what normalization do we aim to. It stinks to periodic potential or periodic configuration space, as a first conjecture.

 

[CarlB]

Yep, when the packet halfwidth is taken as 1, the wavenumber is k0= i pi, or perhaps i pi/4 depending on what normalization do we aim to. It stinks to periodic potential or periodic configuration space, as a first conjecture.

Or perhaps one hidden dimension, cyclic, which is what my Clifford algebra density matrix analysis of the situation requires. That is, using primitive idempotents (projection operators or pure density matrices) you can count the hidden dimensions of spacetime by looking at the weak hypercharge and weak isospin quantum numbers:
http://brannenworks.com/a_fer.pdf

With this sort of idea, the arbitrary complex phase universal to all quantum mechanics becomes a position coordinate in the hidden dimension.

 

[arivero]

In the spirit of keeping this thread for numbers and leaving (most of) the standard theory in other hands, I have hitchhiked to the thread

https://www.physicsforums.com/showthread.php?t=240247

to discuss on EM duality and like. Here, let me just to note that a fix of notation, from Dirac 1948:

e_0^2={1 \over 137} \hbar \ c
g_0^2=4 {137 \over 1} \hbar \ c

for n=1 in {1 \over 2} n \hbar \ c

The point is that there was other papers where it is argued that n must be even. Let me call g_2 to this forceful even constant. In this case
g_2^2= {137 \over 1} \hbar \ c

As for the 4 pi factor, it seems, looking at Dirac's paper, that it was because some other publications use h instead of \hbar, and this use has propagated until today.

I need to locate the paper where it is argued that n=2 is the minimum case.

 

[arivero]

In the spirit of keeping this thread for numbers and leaving (most of) the standard theory in other hands, I have hitchhiked to the thread

https://www.physicsforums.com/showthread.php?t=240247

to discuss on EM duality and like. Here, let me just to note that a fix of notation, from Dirac 1948:

e_0^2={1 \over 137} \hbar c = \alpha_e \hbar \ c
g_0^2=4 {137 \over 1} \hbar c= 2^2 {1 \over \alpha_e} \hbar \ c

for n=1 in
eg= {1 \over 2} n \hbar c

The point is that there was other papers where it is argued that n must be even. Let me call g_2 to this forceful even constant. In this case
g_2^2= {137 \over 1} \hbar \ c

As for the 4 pi factor, it seems, looking at Dirac's paper, that it was because some other publications use h instead of \hbar, and this use has propagated until today.

I need to locate the paper where it is argued that n=2 is the minimum case.

 

[Hans de Vries]

In the spirit of keeping this thread for numbers and leaving (most of) the standard theory in other hands, I have hitchhiked to the thread

https://www.physicsforums.com/showthread.php?t=240247

to discuss on EM duality and like. Here

One can of course consider the continuous spin-density distribution of an electron
field as a distribution of parallel Dirac strings. (Which are basically solenoids)
In one way or another this could lead to charge quantization in the Dirac sense.

Jackson, in section 6.12 mentiones this issue of n versus n/2. There are semi-classical
derivations from Saha (1936) and Wilson (1949) of the Dirac condition which lead to n.


Regards, Hans

 

[arivero]

One can of course consider the continuous spin-density distribution of an electron
field as a distribution of parallel Dirac strings. (Which are basically solenoids)
In one way or another this could lead to charge quantization in the Dirac sense.

It seems that the interaction between "Dirac-Schwinger-Zwinger-Winger" quantization and topological solutions of electromagnetism is a well known candidate to fix the fine structure constant. Ketov 9611209v3 starts his lecures on Seiberg-Witten underlining that "the sole existence of duality symmetry allows one to exactly determine the critical temperature which must occur as the self-dual point where K=K* or sinh(2J/k T)=1". And, more in our electromagnetic constext, Julia and Zee 1975 tell that

If for some reason ... can only take on discrete values ... and if the argument of Schwinger and Zwanziger is relevant for the present case, the one would apparently be faced with a misterious condition saying that the theory will only make sense for some definite value of e^2

 

[Hans de Vries]

It seems that the interaction between "Dirac-Schwinger-Zwinger-Winger" quantization and topological solutions of electromagnetism is a well known candidate to fix the fine structure constant. Ketov 9611209v3 starts his lecures on Seiberg-Witten underlining that "the sole existence of duality symmetry allows one to exactly determine the critical temperature which must occur as the self-dual point where K=K* or sinh(2J/k T)=1". And, more in our electromagnetic constext, Julia and Zee 1975 tell that

Stimulating to follow this line are Tonomura's video's of real life toplological solutions.

http://www.hqrd.hitachi.co.jp/global/movie.cfm
https://www.amazon.com/dp/9810225105/?tag=pfamazon01-20

Specially movie no. 5 with shows the annihilation of "particle/ anti-particles" (vertices/anti-
vertices) There's a close relation of these vertices with the Dirac's charge quantization.

Concerning introductions in the major QFT textbooks, there's Ryder and more recently
Zee for those interested in the general topic.


Regards, Hans

 

[arivero]

Damn, it is alpha=1

It just dawned on me: the square root of 4 pi is not a normalisation factor nor a volume factor; it is the charge of the electron when the fine structure constant is unity.

So Hans formula is

<br /> e_{\alpha=1/137...} + g_{\alpha=1/137...} = e_{\alpha=1} \ {\bf e}^{\pi^2 \over 4}<br />

where
e_{\alpha=1/137...}= \sqrt{4 \pi \alpha} is the electric charge of the electron
g_{\alpha=1/137...}= 4 \pi / e_{\alpha=1/137...} is the DSZ charge of the monopole.
e_{\alpha=1}= \sqrt{4 \pi} the charge of the electron in the fixed point of duality (alpha=1)

Note that if we restore units h,c, they appear congruently in the three terms, because the 4pi of DSZ inversion is actually 4pi hbar c, or 2hc

As for the scaling/rotation factor, the two best ansatzes I have got are still
{\bf e}^{\pi^2 \over 4} = (\sinh {\pi^2 \over 4} + \cosh {\pi^2 \over 4} )
and
\sqrt {4 \pi} \ {\bf e}^{\pi^2 \over 4} = \int {\bf e}^{\pm \frac 12 \pi x } {\bf e}^{-{x^2 \over 4}} dx = k \int_{-\infty}^{\infty} {\bf e}^{\pm \frac 12 k \pi x} {\bf e}^{-{k^2 x^2 \over 4}} dx for any k.

the idea of the second one is that the integral is the square modulus of an Euclidean wavepacket (thus of imaginary wavenumber i k pi). Wick rotating back to Minkowski, the wavenumber gets real, its exponential regains the usual imaginary factor, and then its "Minkoswki" square modulus is
k \int_{-\infty}^{\infty} {\bf e}^{-{k^2 x^2 \over 4}} dx = \sqrt {4 \pi}

 

[arivero]

it is h, not pi!

Er did I say "natural units". Of course in natural units 1=hbar = h/ 2 pi. So a new conjecture for the formula is

<br /> <br /> e_{\alpha=1/137...} + g_{\alpha=1/137...} = e_{\alpha=1} \ {\bf e}^{h^2 \over 16}<br /> <br />

It has a satisfactory point, all the objects are now quantum objects. But it has a couple problems: it is not clear how units are rebuilt in the exponent, and the 16 is a pity; I had hoped it to disappear too :-(.

In any case, the point is that the pi is not coming from the compactification of a KK circle: it is coming directly from Heisenberg principle when applied to the wavepacket. The wavepacket is actually a regularisation family for a dirac delta; one could even hope that in the cero radius limit the exponential factor could be the same for every regularisation; it is perhaps too much to ask for.

 

[CarlB]

For some reason, the semiclassical approximation to QCD shown in this paper seemed to m to be germane to the alpha calculation:


Yang-Mills Propagators and QCD
Marco Frasca
We present a strong coupling expansion that permits to develop analysis of quantum field theory in the infrared limit. Application to a quartic massless scalar field gives a massive spectrum and the propagator in this regime. We extend the approach to a pure Yang-Mills theory obtaining analogous results. The gluon propagator is compared satisfactorily with lattice results and similarly for the spectrum. Comparison with experimental low energy spectrum of QCD supports the view that  resonance is indeed a glueball. The gluon propagator we obtained is finally used to formulate a low energy Lagrangian for QCD that reduces to a Nambu-Jona-Lasinio model with all the parameters fixed by those of the full theory.
http://arxiv.org/abs/0807.4299

 

[Hans de Vries]

at least the new 2008 value for alpha from gabrielse/kinogarbagea begins to
solidify the n=3 term which now gives some convidence in the whole series.

\alpha\ =\ 1/137.035999084 (51)

the alpha "radiative" series:

\begin{aligned}&amp;. \\ &amp;.~~~~\sum_{n=0}^\infty~\frac{\mbox{\huge $\alpha$}^{n-\frac{1}{2}}}{(2\pi)^{b_n}} ~~ = ~~ \mbox{\huge $ e^{\frac{~\pi^2}{4}}$}~~~~. \\ &amp;. \end{aligned}

where b_n is the binominal series 0,0,1,3,6,10.. With successive increments
of 0,1,2,3,4...

apparently somebody solved it for 1294 digits

http://science6.2ch.net/test/read.cgi/sci/1091534329/l50

[size="2"]
1/α=
137.0359990958 2970048964 7400982482 4649832472 5408221072 8280453419 8236353775 3291508251 5164063679 7696140100
    0562638972 3955307219 1131054898 8411966378 9253597119 3431450591 9378868411 2253831430 9814642191 1084940907
    8132259876 7066274120 1689553474 9369387943 1203244561 3909631074 6327549332 6111965801 5656960227 1639347243
    0757007082 5600347556 4972841179 8177526459 2238581338 1031279567 7161638183 9778886042 0778973428 2353736066
    0555206708 9800179254 4185014141 4229664797 4602586290 4504613346 3047999465 7523213673 2504266689 6696242124
    0527823171 5115527412 2534727726 0326504468 2081772605 6146751666 7986300114 2671896058 3800062750 9682551272
    0001496187 4633319304 9496438649 4087857886 0022706370 1852275955 5736144810 3351280327 2115723542 9277212344
    6351827920 3207784574 9582376454 9389618767 5666772174 2641243315 1604981948 7898184100 7657369800 7939844563
    0773213335 5609920096 1118610374 0331148902 0684994710 0221083527 2203073949 1900453584 3300826440 5980415900
    4846789648 2011040441 7960811223 2721740912 7687806462 3818793279 0079147864 9620502089 5698441703 6636685100
    2052169799 1019507098 3686916619 8730033698 5715452876 0094017710 1259020172 1556975359 8282754919 5804007348
    2935098701 8839649369 8755374879 7396746681 2007580127 6048718855 2240202268 0518075853 2654287399 3572658304
    7624575104 9403524357 1121045001 8139733681 5436362649 7016835615 5476863905 7693063483 1562871355 1･･･ [/size]

Regards, Hans

 

[granpa]

from post #342:

where the Koide formla is more convincing is that it is consistent with exactly three generations and no more. The short form for the Koide formula is:

\sqrt{m_n} = 1 + \sqrt{2}\cos(2n\pi/3 + 2/9 + \epsilon)

where \epsilon = 0.22222204717(48) - 2/9 and I've left off an overall scaling factor. Since 2(n+3m)\pi/3 = 2n\pi/3 + 2m\pi, the formula gives exactly three masses so there are only three generations implied. These are the electron, muon, and tau for n the generation number, 1, 2, 3. The \sqrt{2} is what Koide found in 1981, the \epsilon is what I found a year ago.
.

it allows 4 generations if you count m=0

 

[Vanadium 50]

Wow...that post you are responding to is two years old.

There is no massless 4th generation of leptons.

 

[granpa]

just in case it wasnt clear here is the koide formula

zero can obviously be added to both numerator and denominator without changing anything.

http://en.wikipedia.org/wiki/Koide_formula

I don't know if that changes anything or not. the math in the rest of this thread is far beyond me.

 

[arivero]

Wow...that post you are responding to is two years old.

There is no massless 4th generation of leptons.

Still, from time to time I wonder if there is a massless 3rd generation of leptons. Meaning that if the electron mass is put equal to zero, the masses of muon and tau still are near of Koide's conditions. On the other hand, in the limit where all the e,u,d generation is massless, the pion should be, and it strikes me: in this limit, it should have the same mass that the electron. In the actual, broken situation, it has almost the same mass that the muon. Besides, if it had exactly the same mass that the muon, the charged pion would be an stable particle.